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Math Help - Mutually Exclusive Events

  1. #1
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    Mutually Exclusive Events

    Hello people, i need some help

    This is the Question:

    Check the truth of the following statement: if P(A)=2/3 P(B) , P(B)=3/8 P(C) and P(C) = 2/3, then the events are mutually exclusive per two. To be honest, i cant understand that "per two" phrase.

    So , two events in order to be mutual exclusive, A /\ B = 0 , P(A\/B) = P(A) + P(B)

    I am not sure how to begin solving this. i noticed also that P(A) + P(B) + P(C) > 1 (13/12).

    Thanks in advance ;-)
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  2. #2
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    Re: Mutually Exclusive Events

    Quote Originally Posted by primeimplicant View Post
    This is the Question:
    Check the truth of the following statement: if P(A)=2/3 P(B) , P(B)=3/8 P(C) and P(C) = 2/3, then the events are mutually exclusive per two. To be honest, i cant understand that "per two" phrase.
    So , two events in order to be mutual exclusive, A /\ B = 0 , P(A\/B) = P(A) + P(B)
    I am not sure how to begin solving this. i noticed also that P(A) + P(B) + P(C) > 1 (13/12).
    You ought to know that:
    1\ge\mathcal{P}(A\cup B\cup C).
    What if A\cap B=\emptyset,~A\cap C=\emptyset,~\&~B\cap C=\emptyset~?
    Thanks from primeimplicant
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    Re: Mutually Exclusive Events

    Quote Originally Posted by Plato View Post
    You ought to know that:
    1\ge\mathcal{P}(A\cup B\cup C).
    Hey Plato, thanks for your response ;-)

    I know that, apparently every probabillity must be <=1
    Quote Originally Posted by Plato
    What if A\cap B=\emptyset,~A\cap C=\emptyset,~\&~B\cap C=\emptyset~?
    Err, that means that the above are mutually exclusive.

    But, how am i supposed to prove this nice? What steps should i take ?
    Last edited by primeimplicant; March 10th 2012 at 01:50 PM.
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    Re: Mutually Exclusive Events

    Quote Originally Posted by primeimplicant View Post
    that means that the above are mutually exclusive.
    But, how am i supposed to prove this nice? What steps should i take ?
    This is a basic axiom of probability.
    If \left\{ {{A_n}} \right\} is any collection of pair-wise disjoint evednts then
    \mathcal{P}\left( {\bigcup\limits_n {{A_n}} } \right) = \sum\limits_n {\mathcal{P}\left( {{A_n}} \right)}  \le 1
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    Re: Mutually Exclusive Events

    Quote Originally Posted by Plato View Post
    This is a basic axiom of probability.
    If \left\{ {{A_n}} \right\} is any collection of pair-wise disjoint evednts then
    \mathcal{P}\left( {\bigcup\limits_n {{A_n}} } \right) = \sum\limits_n {\mathcal{P}\left( {{A_n}} \right)}  \le 1
    Ok cool,

    After some thought....

    {P}(A\cup B\cup C) = P(A)+P(B)+P(C) - P(A/\B) -P(A/\C)-P(B/\C) +P(A/\B/\C)

    So, if the events are mutually exclusive, then P(A/\B)=P(A/\C)=P(B/\C)=P(A/\B/\C) =\emptyset.~ Is it alright up to this point ?

    Then, i conclude that : {P}(A\cup B\cup C) = P(A)+P(B)+P(C) .

    BUT P(A)+P(B)+P(C) = 13/12 . So if my calculation is right, P(A)+P(B)+P(C) > 1 implies that {P}(A\cup B\cup C) > 1 . Thats against the basic axiom of probability.

    Thus, there is no truth on the given statement.

    Its this right, or i am missing something?

    Thanks!!
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    Re: Mutually Exclusive Events

    Quote Originally Posted by primeimplicant View Post
    then, i conclude that : {p}(a\cup b\cup c) = p(a)+p(b)+p(c) .
    But p(a)+p(b)+p(c) = 13/12 . So if my calculation is right, p(a)+p(b)+p(c) > 1 implies that {p}(a\cup b\cup c) > 1 . Thats against the basic axiom of probability.
    Thus, there is no truth on the given statement.
    correct!
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    Re: Mutually Exclusive Events

    Quote Originally Posted by Plato View Post
    correct!
    Kudos!! That was pretty basic i guess, eh ? I am so beginner lol ... Thanks again pal
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